LeetCode题解： Construct Binary Tree from Inorder and Postorder Traversal

Construct Binary Tree from Inorder and Postorder Traversal

Given inorder and postorder traversal of a tree, construct the binary tree.

Note:

You may assume that duplicates do not exist in the tree.

思路：

已知中序遍历和后序遍历的结果，要求反推树结构。首先注意到后序遍历中，最后一个元素必然是树的根结点。因为树的值没有重复，所以可以在中序遍历中寻找对应的值，这个值左右部分分别是根结点的左子树和右子树，从而知道左右子树元素的个数和中序遍历的结果。再反过来，因为后序遍历首先遍历左子树，再遍历右子树，所以同样可以得到后序遍历的结果。递归构造即可得到原始树结构。

题解：

/**
* Definition for binary tree
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
TreeNode* buildTree(const vector<int>& inorder, int is, int ie,
const vector<int>& postorder, int ps, int pe)
{
if (is > ie)
return nullptr;

int base = postorder[pe];
TreeNode* node = new TreeNode(base);

int ipos = is;
while(inorder[ipos] != base) ++ipos;

int left_nodes = ipos - is;
int right_nodes = ie - ipos;

if (left_nodes != 0)
node->left = buildTree(inorder, is, is + left_nodes - 1,
postorder, ps, ps + left_nodes - 1);

if (right_nodes != 0)
node->right = buildTree(inorder, ipos + 1, ie,
postorder, ps + left_nodes, pe - 1);

return node;
}

TreeNode *buildTree(const vector<int> &inorder, const vector<int> &postorder) {
if (inorder.empty())
return nullptr;

return buildTree(inorder, 0, inorder.size() - 1,
postorder, 0, postorder.size() - 1);
}
};